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Energy problem with friction

  1. Sep 1, 2016 #1
    1. The problem statement, all variables and given/known data
    The brick mass 500g is pushed on hill with velocity 200 cm/s, how far can the brick reach before it stops when μ=0.150
    Untitled.png
    2. Relevant equations
    W=Fdcosθ
    E=1/2mvv
    3. The attempt at a solution
    I got about 1 m before the brick stops, does it correct?
     
  2. jcsd
  3. Sep 1, 2016 #2

    PeroK

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    How did you calculate 1m?
     
  4. Sep 1, 2016 #3
    I calculated how much energy in the brick after it was pushed with Kinetic equation and I got 1 J, when the brick stops moving it means the energy was consumed by the friction till 0J, I can find work by W=(delta)E and I got 0-1= -1 J

    -1=Fdcosθ
    F is friction force = 0.735 N
    d is the displacement
    cosθ is anagle between F and d (180 degree) = -1
    -1=(0.735)(d)(-1)
    it is about 1m (significant digit has only 1 )
     
  5. Sep 1, 2016 #4

    PeroK

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    Aren't you forgetting something? Hint: What would happen if friction were zero?
     
  6. Sep 1, 2016 #5

    jbriggs444

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    How did you calculate F? How should you have calculated F?
     
  7. Sep 1, 2016 #6
    I gave up ;_; there are so many variables to thing about. I should go back and re-study again
     
  8. Sep 1, 2016 #7

    jbriggs444

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    The way to calculate F is to start by calculating the normal force of the 500g block on the ramp. A free body diagram would allow you to clearly see what forces act on that block. Concentrate on the [component of] the forces in the direction perpendicular to the ramp. There are two. The block does not accelerate perpendicular to the ramp. That gives you an easy equation to solve for the normal force.
     
  9. Sep 1, 2016 #8
    OK, I missed something, I forgot that the only force acting on the object is friction
    W=Fdcosθ
    E(final) - E(intial) = F(friction)d(displacement)cos180
    mgh- 1/2mv2 = μNd(-1)

    (0.5)(-9.8)(dsin25)-1/2(0.5)(2)2 = (0.150)(0.5)(-9.8)(d)(-1)
    -2d-1 = 0.735d
    d = -1 / (2-0.735)
    d ≈ -0.8 m (opposite direction of friction )

    Does it correct?
     
  10. Sep 2, 2016 #9

    haruspex

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    Do you expect the left hand side of that equation to turn out positive or negative?
    Does that have the right sign for the left hand side?
    Has the brick gained or lost PE?
     
  11. Sep 2, 2016 #10
    oppps I forgot the negative

    what do you mean about the brick gained or lost PE? The brick loses some of the energy by the work of friction?
     
  12. Sep 2, 2016 #11

    jbriggs444

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    That is not correct. There are two other forces acting on the object.

    You need to correct your calculation of F. It is not given by 0.5 times -9.8. Nor is is clear why you quote cosine in one formula and then sine in the next.

    Edit: possibly you slipped from calculating friction to calculating GPE and did not mention it.
     
  13. Sep 2, 2016 #12
    There are normal force and gravity too
    Can I find Work by minus the final energy(potential) with inital energy(kinetic)

    Solving this problem for 3 days really make me feel bad about myself but I will feel worse if just leaving it like this :/
     
  14. Sep 2, 2016 #13

    jbriggs444

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    Right you are. And yes, the work done by friction plus the work done by the normal force will be equal to the change in (potential plus kinetic) energy. You can ignore the work done by the force of gravity as long as you are already accounting for it as gravitational potential energy.
     
  15. Sep 2, 2016 #14

    haruspex

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    I read your term (0.5)(-9.8)(dsin25) as being the change in PE, but the sign is wrong.
    Because of the above, I presumed that had been a typo, a missing 'not'.
     
  16. Sep 2, 2016 #15

    jbriggs444

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    @haruspex, I was thinking of the normal force. Irrelevant to work done, but for completeness...
     
  17. Sep 2, 2016 #16

    haruspex

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    Yes, I realised that just after posting, wasn't quick enough in my editing.
     
  18. Sep 2, 2016 #17
    sss.png
    Did I miss something? I guess I will never find answer until I can draw correctly
     
  19. Sep 2, 2016 #18

    haruspex

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    That is correct, as long as you bear in mind that the mg sin(25) and mg cos(25) terms replace the mg term. They should not really all be shown in the same diagram.
     
  20. Sep 2, 2016 #19
    here is my final equation

    PE - KE = Fdcosθ
    mgh - 1/2mv2 = ( μmgcos25+mgsin25 ) d cos180
    gh - 1/2v2 = [ μg(0.9) + g(0.42) ] d (-1)
    (-9.8)(dsin25) - 1/2(2)2 = [ (0.150)(-9.8)(0.9) + (-9.8)(0.42) ] d (-1)
    (-9.8)(d)(0.42) - 2 = [ -1.323 + -4.116 ] d (-1)
    -4.14d-2 = 5.44d
    d = -2/ (5.44+4.14)
    d = -0.20 m


    and again the displacement is negative it means the direction of the displacement is opposite of the friction , What if I want displacement to be positive?
     
  21. Sep 2, 2016 #20

    haruspex

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    You still have not taken into account my comment about the sign in post #14. If you are measuring d as positive up the slope (which you appear to be doing by putting a cos(180) on the right hand side of the equation) then your expression for the gain in PE is negative. That is obviously not right.
    m(-9.8)(dsin25) is the work done by gravity. The gain in PE is minus that.
    It is so easy to get the wrong signs in these equations that it is essential to do a sanity check each time. Did it make sense here that the gain in PE is negative?
     
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